Page 118 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 11: Rouche’s Theorem

z 2 z Notes
1 z   ...  n , has modulus > R. Or to put it another way, given an R there is an N so that for

2 n!
z 2 z
n > N no polynomial 1 z   ...  n , has a zero inside the circle of radius R.
2 n!
11.3 Summary

Let C be a simple closed curve, and suppose f is analytic on C. Suppose moreover that the

only singularities of f inside C are poles. If f(z)  0 for all z  C, then  = (C) is a closed curve
which does not pass through the origin. If
(t),   t  
is a complex description of . Now, let’s compute

f'(z)  f'( (t))

 f(z) dz   f( (t))  '(t)dt.
C  
But notice that ’(t) = f’((t))’(t). Hence,

f'(z)  f'( (t))   '(t)

 f(z) dz   f( (t))  '(t)dt   (t) dt
C    
where |n| is the number of times  ”winds around” the origin. The integer n is positive
in case  is traversed in the positive direction, and negative in case the traversal is in the
negative direction.

f'(z)
We shall use the Residue Theorem to evaluate the integral dz. The singularities of
  f(z)
C
f'(z)
the integrand are the poles of f together with the zeros of f. Let’s find the residues at
f(z)
these points. First, let Z = {z , z , ..., z } be set of all zeros of f. Suppose the order of the zero
K
1
2
z is n. Then f(z) = (z – z) j h(z) and h(z)  0. Thus,
n
j
j
j
j
f'(z) (z p ) h'(z) m (z p ) m j 1 h(z) (z p )m j
m



j

j
j
j
j
f(z) = (z pj) 2m j . h(z)

h'(z) m
=  j .
h(z) (z p ) m j

j
Suppose f and g are analytic on and inside a simple closed contour C. Suppose that |f(z)|

> |g(z)| for all z  C. Then we shall see that f and f + g have the same number of zeros
inside C. This result is Rouche’s Theorem. To see why it is so, start by defining the function
(t) on the interval 0  t  1 :
1 f'(z) tg'(t)

 (t)   dz.
2 i C f(z) tg(z)


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